Question

Exponential equation-
Find exact solution and decimal approx.
49x+7^x+1-30=0

Answer 1

Is it
49^x+7^(x+1)-30=0

Answer 2

For decimal approx: I got 0.565

Answer 3

solution: log(3)/log(7)

Answer 4

\[49^x+7^{x+1}-30=0\]
\[(7^{x})^{2}+7^{x+1}-30=0\]
\[(7^{x})^{2}+7^{x}*7^{1}-30=0\]
Let 7^x= y
\[y^2+7y-30=0\]
\[(y+10)(y-3)=0\]
\[y=-10,3\]
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\[7^x=-10~~,~7^x=3\]
\[\log _{7}(-10)=x~~,~~\log _{7}3=x\]

Answer 5

\[\log _{7}3=x\]
\[x=\frac{\log_{}3}{\log7}\]